Conductivity of $0.001 \ M$ aqueous solution of $Na_2SO_4$ is found to be $2.6 \times 10^{-3} \ S \ cm^{-1}$ at $25 \ ^oC$. If limiting molar conductance of $Na^+$ is $50 \ S \ cm^2 \ mol^{-1}$,then limiting molar conductance of $SO_4^{2-}$ will be .............. $S \ cm^2 \ mol^{-1}$ (neglect conductivity of water).

The molar conductivity of a $0.5 \ mol/dm^3$ solution of $AgNO_3$ with electrolytic conductivity of $5.76 \times 10^{-3} \ S \ cm^{-1}$ at $298 \ K$ is ......... $S \ cm^2/mol$.

Resistance of a cell containing $0.02 \ M \ KCl$ solution is $164 \ \Omega$. If the cell is filled with $0.05 \ M \ AgNO_3$,the resistance becomes $75.8 \ \Omega$. Calculate the following: [Conductivity of $0.02 \ M \ KCl = 2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$] $(i)$ Conductivity of $0.05 \ M \ AgNO_3$ (ii) Molar conductivity of $AgNO_3$ solution.

$A$ conductivity cell has two electrodes $18 \ mm$ apart and having cross-sectional area $2.0 \ cm^2$. What is the value of the cell constant (in $cm^{-1}$)?

$A$ $KCl$ solution of conductivity $0.14 \, S m^{-1}$ shows a resistance of $4.19 \, \Omega$ in a conductivity cell. If the same cell is filled with an $HCl$ solution,the resistance drops to $1.03 \, \Omega$. The conductivity of the $HCl$ solution is $....... \, \times 10^{-2} \, S m^{-1}$. (Round off to the Nearest Integer).

The resistance of a conductivity cell filled with $0.1 \ M$ electrolyte solution is $100 \ \Omega$. The conductivity of this solution is $1.29 \ S \ m^{-1}$. If the resistance of the same cell filled with $0.02 \ M$ solution is $520 \ \Omega$,what is the molar conductivity of the $0.02 \ M$ solution?